Advertisements
Advertisements
प्रश्न
Solve the following equation by factorisation :
`sqrt(x + 15) = x + 3`
Advertisements
उत्तर
`sqrt(x + 15) = x + 3`
Squaring on both sides
x + 15 = (x + 3)2
⇒ x2 + 6x + 9 – x – 15 = 0
⇒ x2 + 5x – 6 = 0
⇒ x2 + 6x – x – 6 = 0
⇒ x(x + 6) –1(x + 6) = 0
⇒ (x + 6)(x – 1) = 0
Either x + 6 = 0,
then x = -6
or
x – 1 = 0,
then x = 1
∴ x = –6, 1
Check :
(i) If x = –6 then
L.H.S. = `sqrt(x + 15)`
= `sqrt(-6 + 15)`
= `sqrt(9)`
= 3
R.H.S. = x + 3
= –6 + 3
= –3
∵ L.H.S. ≠ R.H.S.
∴ x = –6 is not a root
(ii) If x = 1, then
L.H.S. - `sqrt(x + 15)`
= `sqrt(1 + 15)`
= `sqrt(16)`
= 4
R.H.S. = x + 3
= 1 + 3
= 4
∵ L.H.S. = R.H.S.
∴ x = 1 is a root of this equation
Hence x = 1.
APPEARS IN
संबंधित प्रश्न
A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.
Solve the following quadratic equations by factorization:
`3x^2 - 2sqrt6x + 2 = 0`
Ashu is x years old while his mother Mrs Veena is x2 years old. Five years hence Mrs Veena will be three times old as Ashu. Find their present ages.
Solve 2x2 – 9x + 10 =0; when x ∈ Q
If sin α and cos α are the roots of the equations ax2 + bx + c = 0, then b2 =
Solve the following equation: `x^2 + (a + 1/a)x + 1 = 0`
There is a square field whose side is 44m. A square flower bed is prepared in its centre leaving a gravel path all round the flower bed. The total cost of laying the flower bed and graving the path at Rs 2. 75 and Rs. 1.5 per square metre, respectively, is Rs 4,904. Find the width of the gravel path.
A two digit number is such that the product of the digits is 12. When 36 is added to this number the digits interchange their places. Determine the number.
Harish made a rectangular garden, with its length 5 metres more than its width. The next year, he increased the length by 3 metres and decreased the width by 2 metres. If the area of the second garden was 119 sq m, was the second garden larger or smaller ?
The hypotenuse of grassy land in the shape of a right triangle is 1 metre more than twice the shortest side. If the third side is 7 metres more than the shortest side, find the sides of the grassy land.
